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2022 AMC 12B Problems/Problem 8

Revision as of 22:19, 13 July 2024 by Mathman239458 (talk | contribs) (I added my own solution which is simply factoring the equation to get the equations of a circle and a hyperbola.)

Problem

What is the graph of $y^4+1=x^4+2y^2$ in the coordinate plane?

$\textbf{(A) } \text{two intersecting parabolas} \qquad \textbf{(B) } \text{two nonintersecting parabolas} \qquad \textbf{(C) } \text{two intersecting circles} \qquad \\\\ \textbf{(D) } \text{a circle and a hyperbola} \qquad \textbf{(E) } \text{a circle and two parabolas}$

Solution 1

Since the equation has even powers of $x$ and $y$, let $y'=y^2$ and $x' = x^2$. Then $y'^2 + 1 = x'^2 + 2y'$. Rearranging gives $y'^2 - 2y' + 1 = x'^2$, or $(y'-1)^2=x'^2$. There are two cases: $y' \leq 1$ or $y' > 1$.

If $y' \leq 1$, taking the square root of both sides gives $1 - y' = x'$, and rearranging gives $x' + y' = 1$. Substituting back in $x'=x^2$ and $y'=y^2$ gives us $x^2+y^2=1$, the equation for a circle.

Similarly, if $y' > 1$, we take the square root of both sides to get $y' - 1 = x'$, or $y' - x' = 1$, which is equivalent to $y^2 - x^2 = 1$, a hyperbola.

Hence, our answer is $\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}$.

~Bxiao31415

Solution 2 (Factoring)

We can subtract $2y^2$ from both sides and factor the left side to get $(y^2-1)^2 = x^4$. If we take the square root of both sides, we are left with two equations: $y^2-1 = x^2$ and $y^2-1 = -x^2$. In the former case, we get $y^2-x^2=1$, which, is the formula for a hyperbola. This means that a hyperbola will appear in the graph of the equation. In the latter case, we get $x^2+y^2=1$, which is the equation of a circle which will also appear in the graph. Looking at our options, this is the only valid answer. Hence, the answer is $\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}$.

~mathman239458


Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

See also

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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